Homotopy equivalence upper triangular matrices and torus

Your idea for (a) is nice, but how can be sure that your maps $f, g$ are continuous? Here is an alternative definition: $$f\left( \begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=(a/\lvert a \rvert,c/\lvert c \rvert) \quad g(v,w)=\begin{bmatrix}v & 0 \\ 0 & w\end{bmatrix}.$$ Then $f \circ g = id_T$ and

$$(g \circ f) \left( \begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=\begin{bmatrix}a/\lvert a \rvert&0\\0& c/\lvert c \rvert\end{bmatrix} .$$ A homotopy $H :g \circ f \simeq id_G$ is defined by $$H(\left( \begin{bmatrix}a&b\\0&c\end{bmatrix}, t \right) = \begin{bmatrix}(1-t)a/\lvert a \rvert + ta & tb\\0& (1-t)c/\lvert c \rvert + tc \end{bmatrix} .$$

For (b) consider the map $h = det \circ g : T \to \mathbb C^*$. We have $h(v,w) = vw$. $\pi_1(T,(1,1))$ has two generators given by $\gamma_0(t) = (e^{2\pi it},1)$ and $\gamma_1(t) = (1,e^{2\pi it})$. Hence $(h \circ \gamma_i)(t) = e^{2\pi it}$ which is the generator of $\pi_1(\mathbb C^*,1)$.

Thus your answer for (b) is correct.

For (a) one can pretty explicitly see that it is homotopy equivalent to the torus just by manipulating the spaces involved. Note that your space of matrices is homeomorphic to the space $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}\bigg| \mathbb{C} \ni a,c \neq 0 \right\} \times \mathbb{C}$. This is homotopy equivalent to $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}\bigg| \mathbb{C} \ni a,c \neq 0 \right\}$ since $\mathbb{C}$ is contractible, and $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}| \mathbb{C} \ni a,c \neq 0 \right\}$ is homeomorphic to $\mathbb{C}^\times \times \mathbb{C}^\times$ which is homotopy equivalent to $S^1 \times S^1$.